#2**+10 **

cos^6x = cos^4x subtract cos^4x ffrom both sides

cos^6x - cos^4x = 0 faactor this

cos^4x*(cos^2x - 1) = 0

cos^4x (cosx + 1)(cosx-1) setting each factor to 0, we have

x = 0, pi/2, pi , (3 pi)/2 in the interval [0, 2pi)

More general solutions are x = 0 + (pi)n and x = pi/2 + (pi/2)n where n is an integer

CPhill
Nov 8, 2014

#1**+5 **

$$\\cos^6x=cos^4x\\

$Divide both sides by $\;cos^4x\\

cos^2x=1\\

cosx=\pm1

$At this point I just think about the unit circle$\\

$cos of an angle is given by the x value (not the angle x)$\\

$The x value is 1 at 0 and it is -1 at 180 degrees$\\

So \\

x=180n$ degrees where $ n\in Z\\

or\\

x=n\pi$ radians where $ n\in Z\\$$

Melody
Nov 8, 2014

#2**+10 **

Best Answer

cos^6x = cos^4x subtract cos^4x ffrom both sides

cos^6x - cos^4x = 0 faactor this

cos^4x*(cos^2x - 1) = 0

cos^4x (cosx + 1)(cosx-1) setting each factor to 0, we have

x = 0, pi/2, pi , (3 pi)/2 in the interval [0, 2pi)

More general solutions are x = 0 + (pi)n and x = pi/2 + (pi/2)n where n is an integer

CPhill
Nov 8, 2014

#3**+5 **

Okay I missed some answers - thanks for picking that up chris but your presentation looks strange.

isn't

x = 0 + (pi)n and x = pi/2 + (pi/2)n

really just

$$x=\frac{n\pi}{2}\qquad where \qquad n\in Z$$

???

Melody
Nov 8, 2014